int Edmonds_Karp(int source, int sink, int vertex_num)
{
    int maxFlow = 0;
    while(true) {
        int head, tail, pre_of[UPPER], my_stack[UPPER];
        head = tail = 0;//初始化栈
        my_stack[tail++] = source;//放入源点
        memset(pre_of, -1, sizeof(pre_of));//pre_of[v],v 代表边u v
        while(head < tail){
            int u = my_stack[head++];//取出u点
            for(int v = 1; v <= vertex_num; v ++){
                if(residual[u][v] > 0 && pre_of[v] == -1) {
                    my_stack[tail++] = v;/*入栈*/pre_of[v] = u;/*保存这条边*/
                    if(u == sink)   break;
                }
            }
            if(pre_of[sink] != -1) break;//找到增广路径
        }
        if(pre_of[sink] == -1) break;//没有增广路径
        int min_flow = INF;
        for(int v = sink; v != source; v = pre_of[v]) {
            min_flow = mymin(min_flow, residual[pre_of[v]][v]);
            //对pre保存的路径进行回溯找到最小的流
        }
        maxFlow += min_flow;
        for(int v = sink; v != source; v = pre_of[v]) {//更新流网络
            residual[pre_of[v]][v] -= min_flow;
            residual[v][pre_of[v]] += min_flow;
        }
    }
    return maxFlow;
}

//部分代码
//纯粹的Ford Fulkerson思想
//纯DFS，回溯时更新流，比模板简单、短很多
//效率略低，10ms
int matrix[200+1][200+1],int pre_of[200+1];
int DFS(int i,int v,int min_flow){
    int k;
    if(i == v){
        pre_of[i] = -1;
        return min_flow;
    }
    for(k = 1;k <= v;k++){
        if(pre_of[k] == -1 && matrix[i][k] > 0){
            pre_of[k] = i;
            int t = DFS(k,v,mymin(min_flow,matrix[i][k]));
            if(t != 0){//有一条增广路径
                matrix[k][i] += t;matrix[i][k] -= t;
                pre_of[k] = -1; return t;
            }
        }
    }
    pre_of[i] = -1;
    return 0;
}
//main函数中的主要部分
int e,v,i;
while(scanf("%d%d",&e,&v) != EOF){
    memset(matrix,0,sizeof(matrix));//0表示没有
    memset(pre_of,-1,sizeof(pre_of));
    for(i = 0;i < e;i++){
        int si,ti,ci;scanf("%d%d%d",&si,&ti,&ci);
        matrix[si][ti] += ci;
    }
    int t,ans = 0;
    while((t = DFS(1,v,INF))){memset(pre_of,-1,sizeof(pre_of));ans += t;}
    printf("%d\n",ans);
}